Answer
$\omega_2=1.91~ rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
According to the conservation of angular momentum
$I_{z_1}\omega_1=I_{z_2}\omega_2~~~$[eq(1)]
Now $I_{z_1}=mk_z^2$
$\implies I_{z_1}=(0.75)(0.125)^2=0.0117Kg/m^2$
Similarly
$I_{z_2}=mk_z^2+\frac{1}{2}mr^2$
We plug in the known values to obtain:
$I_{z_2}=(0.75)(0.125)^2+(\frac{0.05}{2})(0.15)^2$
This simplifies to:
$I_{z_2}=0.01228Kg/m^2$
We plug in the known values in eq(1) to obtain:
$(0.0117)(2)=(0.01228)\omega_2$
This simplifies to:
$\omega_2=1.91~ rad/s$
